counting alphabets from file python

Question

read the text from file.txt and count the number of each alphabets in that and out put should be like a=2 ( if there are 2 a's) and b=6 (if there are 6 b's) so far i have done this much, this prints every alphabet but i want to print alphabets which exists only.

f= open('cipher.txt')
word= " ".join(line.strip() for line in f)
word=word.lower()
alpha="abcdefghijklmnopqrstuvwxyz"
alpha=list(alpha)
for i in alpha:
  print(i+"="+str(word.count(i)))

But if any alphabet which is 0 times used i dont want to print that,, how to fix this? Help PLEASE

it doesnt print it gives error, cant change int to str implicitly. — Hasnain Ali, Sep 02 '13 at 08:18
and if the word exists zero time i dont want it to print, how to do that — Hasnain Ali, Sep 02 '13 at 08:24
You could use an `if` statement to check if `word.count(i) > 0`. — korylprince, Sep 02 '13 at 17:16

score 0 · Answer 1 · answered Sep 02 '13 at 08:24

0

import letters
s = 'hello world'
{a:s.count(a) for a in string.letters if s.count(a) != 0}
OUTPUT:
{'d': 1, 'e': 1, 'h': 1, 'l': 3, 'o': 2, 'r': 1, 'w': 1}

answered Sep 02 '13 at 08:24

Denis

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counting alphabets from file python

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